dc123 wrote:
If the integers a and n are greater than 1 and the product of the first 8 positive integers is a multiple of a^n, what is the value of a?
1) a^n = 64
2) n=6
I cant find this question on the site
Such questions try to trick you with an 'obvious c' i.e. you can very easily get the answer using both the statements together. The trick is generally to find the one statement which alone is sufficient (more often that not, there will be one statement which is sufficient in such questions)
In this question we know that a and n are positive integers greater than 1.
Using both the statements together, we can easily find 'a'. But there is a catch.
Let us review each statement independently.
Given: product of the first 8 positive integers is a multiple of a^n
\(1*2*3*4*5*6*7*8 = a^n * k\) (where k is a positive integer)
What does this imply? It implies that every factor of a^n has to be a factor of \(1*2*3*4*5*6*7*8\) as well.
1) \(a^n = 64\)
64 will be a factor of \(1*2*3*4*5*6*7*8\), of course. But can we find 'a' now?
\(2^6 = 64; 4^3 = 64; 8^2 = 64\)
a can take any value 2/4/8. Not sufficient.
2. n = 6
'a' can be 2 since 2^6 = 64 which is a factor of \(1*2*3*4*5*6*7*8 = 64*3*5*6*7\)
Can 'a' be 3?
Is 3^6 a factor of 1*2*
3*4*5*
6*7*8? No, it has only two 3's. It doesn't have six 3's.
Can 'a' be 4?
Is 4^6 a factor of 1*2*3*
4*5*
6*7*
8? No, it has only three 4's.
The same thing will be true for any number greater than 3 too.
Since 'n' is fixed here i.e. 6, 'a' can take only one value i.e. 2
Hence stmnt 2 alone is sufficient. 'a' can only be 2.
Answer (B)